Convert 0.375 to the same 12-bit format.

Denary 0.375 is 0.011 in binary, which normalises to 0.11 × 2⁻¹. The mantissa is 01100000 and the exponent is 1111.

State the value of the floating point number with mantissa 01101000 and exponent 0010.

The mantissa 0.1101 is 0.8125 and the exponent is 2. The value is 0.8125 × 4, which is 3.25.

State the value of the floating point number with mantissa 01000000 and exponent 0101.

The mantissa 0.1 is 0.5 and the exponent is 5. The value is 0.5 × 32, which is 16.

Convert −5.5 to the same 12-bit format.

The mantissa for positive 5.5 is 01011000, whose two's complement is 10101000. The exponent stays 0011, so −5.5 is the mantissa 10101000 with exponent 0011.

Eduqas A-Level Computer Science A500QS

Data Structures: Arrays, Records and Hash Tables

Topic overview One-dimensional arrays and indexing Two- and three-dimensional arrays Manipulating arrays: insertion, deletion and searching Records and fields Arrays of records and record manipulation Hash tables, hashing and collisions Choosing and justifying a data structure

Stacks, Queues, Linked Lists and Trees

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Boolean Logic and Simplification

Topic overview AND, OR and NOT truth tables XOR, NAND and NOR truth tables Truth tables for combined Boolean expressions Boolean identities and rules De Morgan's laws Simplifying Boolean expressions Logical operators in programs, masking and encryption

Algorithms, Pseudo-Code and Programming Constructs

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Variables, Scope, Parameters and Identifiers

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Recursion and Modular Programming

Topic overview Recursion and the base case How recursive calls use the stack Recursive algorithms: factorial and Fibonacci Recursion compared with iteration Integer division with DIV and MOD Standard functions and standard modules User defined subprograms and modular design

Sorting Algorithms

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Searching and Shortest-Path Algorithms

Topic overview Linear search Binary search Comparing linear and binary search Graphs, weights and the shortest-path problem Dijkstra's shortest-path algorithm Tracing a shortest-path algorithm by hand Choosing the right searching technique

Big O Notation and Algorithm Efficiency

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Validation, Verification and Testing

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Data Compression

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Programming Paradigms and Computer Languages

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Systems Analysis, Maintenance and Documentation

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Software Engineering Tools and Translators

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Computing, Society and the Law

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Computer Architecture, the Processor and Assembly Language

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Input, Output and Secondary Storage

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Networks, Protocols and the Internet

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Data Transmission and Network Performance

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Number Systems and Binary Arithmetic

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Data Types, Characters and Floating Point

Topic overview Primitive data types Storage requirements for each data type ASCII and standardised character sets Unicode and multilingual encoding Floating point form: mantissa and exponent Converting between real numbers and floating point Integer and floating point compared Truncation, rounding and accuracy

Files, Records and File Organisation

Topic overview The purpose of files in data processing Files, records and fields Fixed and variable length fields and records Master files and transaction files Serial and sequential file access Indexed sequential access and multi-level indexes Direct access and hashing algorithms Overflow handling and file reorganisation File security, backups, generations and archiving

Databases, Normalisation and Entity Modelling

Topic overview Data consistency, redundancy and independence Relational databases and other database models Primary keys, foreign keys and indexes First normal form Second normal form Third normal form Entity relationship modelling User views and turning data into information

SQL, Big Data and Distributed Systems

Topic overview Query languages and the purpose of SQL Creating tables: CREATE TABLE, data types and keys Retrieving data with SELECT, FROM and WHERE Comparison operators, IN, AND, OR and sub-queries Sorting, grouping and aggregate functions Inserting, updating, deleting and joining data The DBMS, the data dictionary and distributed databases Big Data, data warehousing, data mining and predictive analytics

The Operating System and Its Modes

Topic overview The need for an operating system and the role of the kernel Managing peripherals, processes and memory protection The user interface and the layers between user and hardware Directory structures and file attributes Utility software Batch processing, real time control and real time transaction systems Types of operating system: single-user, multi-user, multi-tasking and multi-programming Designing an interface for different users and environments

Interrupts, Memory Management and Scheduling

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Types of Software System

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Data Security, Cryptography and Attack

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Data Types, Characters and Floating Point

Converting between real numbers and floating point

3 things to know
Component 2

Overview

Turning a real number into a normalised mantissa and exponent and back again, worked in one stated format throughout, with the normalisation rule that fixes the mantissa.

What you need to know

Answer each one before you reveal it.

Your recallNot started

Key term

How is a negative value written in floating point, and how is a mantissa normalised?

Process

Describe the steps for converting a positive real number to normalised floating point.

Numbers

Convert 5.5 to a 12-bit format with an 8-bit mantissa and a 4-bit exponent.

Questions from this subtopic

Each one with the model answer underneath.

1
Describe the steps for converting a positive real number to normalised floating point.
Model answer
The number is first written in binary, then the binary point is moved to sit just after the sign bit. The number of places moved becomes the exponent.
2
Convert 5.5 to a 12-bit format with an 8-bit mantissa and a 4-bit exponent.
Model answer
Denary 5.5 is 101.1 in binary, which normalises to 0.1011 × 2³. The mantissa is 01011000 and the exponent is 0011.
3
Convert 0.375 to the same 12-bit format.
Model answer
Denary 0.375 is 0.011 in binary, which normalises to 0.11 × 2⁻¹. The mantissa is 01100000 and the exponent is 1111.
4
State the value of the floating point number with mantissa 01101000 and exponent 0010.
Model answer
The mantissa 0.1101 is 0.8125 and the exponent is 2. The value is 0.8125 × 4, which is 3.25.
5
State the value of the floating point number with mantissa 01000000 and exponent 0101.
Model answer
The mantissa 0.1 is 0.5 and the exponent is 5. The value is 0.5 × 32, which is 16.
6
Convert −5.5 to the same 12-bit format.
Model answer
The mantissa for positive 5.5 is 01011000, whose two's complement is 10101000. The exponent stays 0011, so −5.5 is the mantissa 10101000 with exponent 0011.

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